Question

A force < mg, −2mg, −3mg > acts on an object as it is

displaced < 2L, 3L, −L/2 >. How much work is done?

Answer 1

The work done by the force on the object is -3mgL.

Step-by-step explanation:

The work done by a force on an object is given by the dot product of the force and displacement vectors. The dot product is calculated by multiplying corresponding components of the two vectors and summing the results.

In this case, the force vector is < mg, -2mg, -3mg > and the displacement vector is < 2L, 3L, -L/2 >. Multiplying the corresponding components and summing the results, the work done is -3mgL.

Therefore, the amount of work done by the force on the object is -3mgL.

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